jmc

By DeMoivre's Theorem, $$\begin{array}{rl}\cos 5\theta +i\sin 5\theta & =(\cos \theta +i\sin \theta {)}^5\\ & ={\cos }^5\theta +5i{\cos }^4\theta \sin \theta -10{\cos }^3\theta {\sin }^2\theta -10i{\cos }^2\theta {\sin }^3\theta +5\cos \theta {\sin }^4\theta +i{\sin }^5\theta .\end{array}$$Equating real parts, we get $$\cos 5\theta ={\cos }^5\theta -10{\cos }^3\theta {\sin }^2\theta +5\cos \theta {\sin }^4\theta .$$Since $\cos \theta =\frac{1}{3},$ ${\sin }^2\theta =1-{\cos }^2\theta =\frac{8}{9}.$ Therefore, $$\begin{array}{rl}\cos 5\theta & ={\cos }^5\theta -10{\cos }^3\theta {\sin }^2\theta +5\cos \theta {\sin }^4\theta \\ & ={\left(\frac{1}{3}\right)}^5-10{\left(\frac{1}{3}\right)}^3\cdot \frac{8}{9}+5\cdot \frac{1}{3}\cdot {\left(\frac{8}{9}\right)}^2\\ & =\boxed{\frac{241}{243}}.\end{array}$$